9. The product of two numbers is 360, If one numnber is l6 less than double the other

Question

number, find the two possible number pairs.

Answer 1

Let the two numbers be \( x \) and \( y \). We have the following relationships:

1. The product of the two numbers: \[ xy = 360 \]

2. One number is 16 less than double the other number. We can express this as: \[ y = 2x - 16 \]

Now, we can substitute the expression for \( y \) into the first equation:

\[ x(2x - 16) = 360 \]

Expanding this gives:

\[ 2x^2 - 16x = 360 \]

Now, rearranging brings all terms to one side:

\[ 2x^2 - 16x - 360 = 0 \]

We can simplify this equation by dividing all terms by 2:

\[ x^2 - 8x - 180 = 0 \]

Next, we will use the quadratic formula, which is given by

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation, \( a = 1 \), \( b = -8 \), and \( c = -180 \). First, we calculate the discriminant:

\[ b^2 - 4ac = (-8)^2 - 4(1)(-180) = 64 + 720 = 784 \]

Now we apply the quadratic formula:

\[ x = \frac{-(-8) \pm \sqrt{784}}{2 \cdot 1} = \frac{8 \pm 28}{2} \]

This gives us two possible solutions for \( x \):

\[ x = \frac{36}{2} = 18 \]

\[ x = \frac{-20}{2} = -10 \]

Now we will find the corresponding values of \( y \) for these \( x \) values:

1. When \( x = 18 \): \[ y = 2(18) - 16 = 36 - 16 = 20 \]

2. When \( x = -10 \): \[ y = 2(-10) - 16 = -20 - 16 = -36 \]

Thus, the two pairs of numbers that multiply to give 360 and satisfy the second condition are:

1. \( (18, 20) \)
2. \( (-10, -36) \)

So, the two possible number pairs are \( (18, 20) \) and \( (-10, -36) \).